C++ Object Pointer

Question

I need some help here, could you tell me how I can get p2 variable content? I can get variables through t1 and p1 but I want to use p2 to get health and mana values.

#include <iostream>

using namespace std;

class Mage{
public:
    int health;
    int mana;
    Mage(int health, int mana){
        this->health = health;
        this->mana = mana;
    }
};

int main(){

    Mage t1 = Mage(1,1);
    Mage *p1 = &t1;
    Mage **p2 = &p1;

    cout << t1.health << endl;
    cout << p1->health << endl;
    cout << "how to print variable content with p2?" << endl;

    return 0;
}

@Beginner has the right of this, but if you find yourself wandering into pointers to pointers outside of a trivial education example, ask yourself "Why am I doing this again?" It's rare that you need more than one level of indirection, and when it crops up, a reference to the pointer (Mage * &p2) is probably a better option. — user4581301
– user4581301, Commented Apr 24, 2017 at 16:29
@user4581301 2 levels of indirection for pointers is rare, yes. But if for example you are iterating over a std::vector of std::unique_ptr, (*it)->foo() needs 2 levels of indirection, there is no other way around it. — Rakete1111
– Rakete1111, Commented Apr 24, 2017 at 16:48

Beginner · Accepted Answer · 2017-04-24 16:35:34Z

2

Use

cout << (*p2)->health << endl;

The * has two meanings: to declare a pointer and to dereference a pointer. But since -> has higher precedence than * you need to put *p2 into backets.

answered Apr 24, 2017 at 16:35

Beginner

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user5420386 · Accepted Answer · 2017-04-24 16:36:09Z

1

Your example suggests that you are looking for a single operator that dereferences two times at once. There is none in C++.

Two ways to achieve what you want. The first has already been mentioned and goes

std::cout << (*p2)->health << std::endl;

The second can be found by noting that p1->health is equivalent to (*p1).health. Hence, (*p2)->health is equivalent to (**p2).health. In a row:

std::cout << t1.health << std::endl;
std::cout << (*p1).health << std::endl;
std::cout << (**p2).health << std::endl;

answered Apr 24, 2017 at 16:36

user5420386

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